I am given a family of problems :
$$ \begin{cases} -\Delta u_\varepsilon = f+\beta_\varepsilon(u_\varepsilon)& \text{in } D \\[2pt] u_\varepsilon=w & \text{on } \partial D \end{cases} $$
with $u_\varepsilon$ solving the above equation in $D\subset \mathbb R^N$ open and bounded.
We have $\beta_\varepsilon(s)=\frac 1 \varepsilon \beta \left( \frac s \varepsilon \right)$.
$\beta$ is a smooth function such that $$ \begin{cases} \beta \geq 0 \\[2pt] \beta \equiv 0 & \text{in } (0,1)^c \\[2pt] \int_{\mathbb R}\beta=1 \end{cases} $$
And $f\in L^\infty(D)$, $0<a<f<A<\infty$, for some $a,A>0$. I wish to prove $u_\varepsilon$ is uniformly bounded in $D$ with bound independent of $\varepsilon$ and $\varepsilon\rightarrow 0$.
The bound may depend on the boundary condition.
It was simple, though the singular term made it look complicated. 1. In the set say $\Omega_{\epsilon}=\{u_\epsilon >\epsilon\}$, we have $$ -\Delta u_{\epsilon}=f\;\;\text {in $\Omega_{\epsilon}$} $$ From Theorem 8.16 in "Gilbarg,Trudinger-Elliptic Partial Differential Equations of Second Order-Springer (2001)" The solution $u_{\epsilon}$ is bounded as long as the boundary term $w$ is bounded,
In rest of the region, $\{u_{\epsilon}\leq \epsilon\}$, we already have a bound.